"How does feedback lower DF"
Magfan It increases DF (lowers output impedance).
Lets take amplifier that has gain of 30 (31.6dB). When input voltage is 1V output voltage is 30V. Output voltage drops (for whatever reason) 1V under 1A load to 29V. That's 1ohm output impedance (DF=8).
Now, let's build this amp with gain of 300 but feed 3% of the output voltage back to the input in opposite phase. As a result amplifiers output is the same 30V as before but input is the difference between 1V and 3% of 30V = 0.1V Lets verify (1V-0.03*30V)*300=30V
Lets load this amplifier with 1A. Our voltage drop inside is still 1V under 1A load, but output voltage will be higher than 29V because we subtract less from the input. Output voltage will be 29.9V and output impedance will be 0.1V/1A=0.1ohm (DF=80). Lets verify. (1V-0.03*29.9V)*300-1Vdrop=29.9V.
Output impedance dropped 10 times. Expression 1+B*Aol is known as Improvement Factor. In our case B (Feedback Factor) = 0.03 (3%), Aol (Open Loop Gain) = 300 thus Improvement Factor = 1+0.03*300=10.
It is a little clumsy, but I didnt want to bring whole feedback theory equations into simple example.